It must be either speeding up, or slowing down, or turning. There are no other possibilities.
The static friction exerted on the block by the incline is 
.
The given parameters;
- <em>mass of the block, = M</em>
- <em>coefficient of static friction in section 1, = </em>
<em /> - <em>angle of inclination of the plane, = θ</em>
<em />
The normal force on the block is calculated as follows;
Fₙ = Mgcosθ
The static friction exerted on the block by the incline is calculated as follows;
Thus, the static friction exerted on the block by the incline is
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Answer:
first blank is chemical second blank is kinetic energy
Answer:
The acceleration is 3.62 m/s²
Explanation:
Step 1: Data given
mass of the shell = 1.65 kg
angle = 38.0 °
Step 2: Calculate the acceleration
We have 2 forces working on the line of motion:
⇒ gravity down the slope = m*g*sinα
⇒ provides the linear acceleration
⇒ friction up the slope = F
⇒ provides the linear acceleration and also the torque about the CoM.
∑F = m*a = m*g*sin(α) - F
I*dω/dt = F*R
The spherical shell with mass m has moment of inertia I=2/3*m*R² Furthermore a pure rolling relates dω/dt and a through a = R dω/dt. So the two equations become
m*a = m*g sin(α) - F
2/3*m*a = F
IF we combine both:
m*a = m*g*sin(α) - 2/3*m*a
1.65a = 1.65*9.81 * sin(38.0) - 2/3 *1.65a
1.65a + 1.1a = 9.9654
2.75a = 9.9654
a = 3.62 m/s²
The acceleration is 3.62 m/s²
